A polynomial may stand beside a root of a ± a cos(y)

BREAKING-CHANGES.md

@@ -1173,6 +1173,22 @@ it in the answer, and what the loop asks is whether the degree has fallen to not
 | `"F^(c*(a+b*x))*sin(d+pe*x)^3".Integrate("x")` | left unevaluated | the antiderivative |
 | `"F^(c*(a+b*x))*sin(d+pe*x)*cos(d+pe*x)".Integrate("x")` | left unevaluated | the antiderivative |
 
+### A polynomial may stand beside a root of `a ± a cos(y)`
+
+`x^3 sqrt(a + a cos(c + d x))` was left as written. The half angle at which `a ± a cos(y)` is a
+square answers the root alone, and it substitutes the angle -- so anything else that mentioned
+the variable stopped it, a polynomial among them. What is left of the variable is the variable
+itself: with `u` half the angle (a quarter turn further for the sine's identity), `x` is
+`(2u - offset)/rate`, so the polynomial is a polynomial in `u`, and what is handed on is one
+times a cosine of `u`, which the closed rules answer. Rubi's 4.2.10
+([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"x^3*sqrt(a+a*cos(c+d*x))".Integrate("x")` | left unevaluated | the antiderivative, with `sgn(cos((c + d x)/2))` in front |
+| `"x^3*sqrt(a-a*cos(x))".Integrate("x")` | left unevaluated | the antiderivative |
+| `"x^2*sqrt(a+a*sin(x))".Integrate("x")` | left unevaluated | the antiderivative |
+
 ### `binomial(n, k)` is a function
 
 **Addition, not silent.** The binomial coefficient is a node, `Entity.Binomialf`, spelled

Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs

@@ -2426,7 +2426,7 @@ Entity Step(ERational power)
                 else if (argument != thisArgument)
                     return null;
             }
-            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out _)
+            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out var offsetOfArgument)
                 || rate.ContainsNode(x) || TreeAnalyzer.IsZero(rate))
                 return null;
             if (!expr.Nodes.All(node => !node.ContainsNode(x)
@@ -2494,8 +2494,19 @@ Entity Step(ERational power)
                 return constant * MathS.Pow(square, Number.Integer.Create(twoP));
             });
             var inU = rewritten.Substitute(sine, sineInU).Substitute(cosine, cosineInU);
+            // What is left of the variable is the variable itself: the angle is `rate x + offset`
+            // and `u` is half of it (a quarter turn on for the sine), so `x` is
+            // `(2u - pi/2 - offset)/rate` there and `(2u - offset)/rate` here -- and
+            // `x^3 sqrt(a + a cos(c + d x))` is a polynomial in `u` times a cosine of it, which
+            // the closed rules answer, where the rule used to decline anything beside the root.
             if (inU.ContainsNode(x))
-                return null;
+            {
+                var argumentInU = sineKind ? (2 * u - MathS.pi / 2).InnerSimplified : (2 * u).InnerSimplified;
+                var xInU = ((argumentInU - offsetOfArgument) / rate).InnerSimplified;
+                inU = inU.Substitute(x, xInU);
+                if (inU.ContainsNode(x))
+                    return null;
+            }
             // dx = 2 du / rate.
             var integrand = (inU * 2 / rate).InnerSimplified;
             if (Integration.ComputeAsAQuestionOfItsOwn(integrand, u, integrateByParts) is not { } result)
@@ -12416,7 +12427,7 @@ bool IsAConstantTimesAPowerOfOneMinusUSquared(Entity polynomial, out Entity cons
                 else if (argument != thisArgument)
                     return null;
             }
-            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out _)
+            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out var offsetOfArgument)
                 || rate.ContainsNode(x) || TreeAnalyzer.IsZero(rate))
                 return null;
             if (!expr.Nodes.All(node => !node.ContainsNode(x)
@@ -14846,7 +14857,7 @@ bool IsAPolynomialInBoth(Entity polynomial)
                 else if (argument != thisArgument)
                     return null;
             }
-            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out _)
+            if (argument is null || !TreeAnalyzer.TryGetPolyLinear(argument, x, out var rate, out var offsetOfArgument)
                 || rate.ContainsNode(x) || TreeAnalyzer.IsZero(rate))
                 return null;
             var sine = MathS.Sin(argument);

Sources/Tests/UnitTests/Calculus/HalfAngleSquareTest.cs

@@ -151,6 +151,26 @@ public void ThePairOfPowersTwoApartIsElementary(string integrand)
             Assert.True(compared >= 4, $"only {compared} of five points were comparable for {integrand}");
         }
 
+        /// <summary>
+        /// A polynomial beside the root: the rule substitutes the half angle, and what is left
+        /// of the variable is the variable itself -- <c>x</c> is <c>(2u - offset)/rate</c> for
+        /// the cosine and a quarter turn further for the sine -- so
+        /// <c>x^3 sqrt(a + a cos(c + d x))</c> is a polynomial in <c>u</c> times a cosine of it,
+        /// which the closed rules answer. It used to decline anything standing beside the root.
+        /// Rubi's 4.2.1. <a href="https://github.com/asc-community/AngouriMath/issues/718">#718</a>
+        /// </summary>
+        [Theory]
+        [InlineData("x^3*sqrt(a + a*cos(c + d*x))")]
+        [InlineData("x*sqrt(a + a*cos(x))")]
+        [InlineData("x^3*sqrt(a - a*cos(x))")]
+        [InlineData("x^2*sqrt(a + a*sin(x))")]
+        [InlineData("x*(a + a*cos(x))^(3/2)")]
+        public void APolynomialMayStandBesideTheRoot(string integrand)
+        {
+            DifferentiatesBack(integrand, 2);
+            DifferentiatesBack(integrand, -3);
+        }
+
         /// <summary>A combination that does not kill the residual is not this rule's, and is left alone.</summary>
         [Fact]
         public void AnotherCoefficientIsLeftAsWritten()